Nested List Weight Sum II

Given a nested list of integers, return the sum of all integers in the list weighted by their depth.

Each element is either an integer, or a list -- whose elements may also be integers or other lists.

Different from the previous question where weight is increasing from root to leaf, now the weight is defined from bottom up. i.e., the leaf level integers have weight 1, and the root level integers have the largest weight.

Example 1:
Given the list [[1,1],2,[1,1]], return 8. (four 1's at depth 1, one 2 at depth 2)

Example 2:
Given the list [1,[4,[6]]], return 17. (one 1 at depth 3, one 4 at depth 2, and one 6 at depth 1; 1*3 + 4*2 + 6*1 = 17) 



The easiest way is to find maximum depth first then calculate sum using the first problem's approach. This requires us to traverse the list twice. A better one pass solution is to keep two variables: weighted and unweighted. When we traverse the list, if we find an integer, we add it to unweighted, otherwise we flatten it and add all its elements to the next level list. Finally we add unweighted to weighted. When we calculate the next level, since the sum of all elements in the upper level is still in unweighted, when we add it to weighted, we add the sum again, which is the depth sum we want.

The approach is actually a BFS approach (similarly to binary tree level order traversal).

public int depthSumInverse(List<nestedinteger> nestedList) {
        int weighted = 0, unweighted = 0;
        while (!nestedList.isEmpty()) {
            List<nestedinteger> nextLevel = new ArrayList<>();
            for (NestedInteger nestedInteger : nestedList) {
                if (nestedInteger.isInteger()) {
                    unweighted += nestedInteger.getInteger();
                } else {
                    nextLevel.addAll(nestedInteger.getList());
                }
            }
            weighted += unweighted;
            nestedList = nextLevel;
        }
        return weighted;
    }